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Chapter 11: Problem 65

You have 50 yards of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?

Short Answer

Expert verified
The dimensions that maximize the area are 12.5 yards by 12.5 yards and the maximum area is 156.25 square yards.

Step by step solution

01

Express the Perimeter in terms of one variable

Since the same amount of fencing (50 yards) is used for both the length and the width of the rectangle, we can express one in terms of the other. Let's define \(l\) as the length and \(w\) as the width. Since the perimeter (P) of a rectangle is given by \(P=2l+2w\), from the given \(P = 50\), we can express width \(w\) in terms of \(l\) as \(w = 25 - l\)
02

Define the Area Function

The area (A) of a rectangle is computed by length times width, that is \(A= l \times w\). Substituting \(w\) from step 1 into the area formula, we get \(A = l \times (25 - l)\), which simplifies to \(A = 25l - l^2\)
03

Maximize the Area

To find the dimensions that maximize the area, take the derivative of the area function, set it equal to zero and solve for \(l\). The derivative of \(A\) with respect to \(l\) is \(A' = 25 - 2l\). Setting this equal to zero, we get \(25 - 2l = 0\), which gives \(l = 12.5\) yards
04

Find the Maximum Area

Substitute \(l = 12.5\) into the area function from step 2 to compute the maximum possible area. Thus, \(A_max = 25 \times 12.5 - (12.5)^2\), which calculates as \(A_max = 156.25\) square yards
05

Find the Width of the Rectangle

We substitute \(l = 12.5\) into our equation from step 1 for \(w\), thus we have \(w = 25 - 12.5 = 12.5\) yards. So the rectangle is a square.

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Most popular questions from this chapter

Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. When I complete the square for the binomial \(x^{2}+b x\) I obtain a different polynomial, but when I solve a quadratic equation by completing the square, I obtain an equation with the same solution set.

Solve by completing the square: $$x^{2}+b x+c=0$$

Solve inequality using a graphing utility. \(\frac{1}{x+1} \leq \frac{2}{x+4}\)

Solve each quadratic equation for \(u\). $$u^{2}-8 u-9=0$$

Will help you prepare for the material covered in the next section. Use point plotting to graph \(f(x)=x^{2}\) and \(g(x)=(x+2)^{2}\) in the same rectangular coordinate system.
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